This deck presents linear maps as the structure-preserving morphisms between vector spaces. It gives the two axioms, explains why the origin is fixed, shows that a basis determines the whole map, and covers the kernel and image as subspaces and injectivity as the kernel being trivial. It targets the classic traps of calling affine or squaring maps linear, thinking that one vector pins a map down, locating the kernel in the wrong space, and assuming that injective forces surjective.
Subject: Foundations of Higher Mathematics · 114 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
A linear map is the right notion of a function between vector spaces: it is exactly the kind of map that respects the two things a vector space is built from.
By the end you can:
1. State the two linearity axioms and test any map against them.
2. Explain why a linear map must fix the origin, and use that to reject affine maps instantly.
3. Use the fact that a linear map is completely determined by what it does to a basis.
4. Compute the kernel and image of a map, prove both are subspaces, and read off injectivity from the kernel.
Warm-up
Discussion prompt
Before we open Linear Maps: without looking back, what was the main idea of Basis & Dimension, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck presents a basis as the fusion of independence and spanning, then covers the unique-coordinate theorem, extending an independent set and shrinking a spanning set, the invariance-of-dimension theorem proved via the exchange lemma, and the Grassmann dimension formula. It targets the classic traps: calling a spanning but dependent set a basis, assuming that any n vectors in an n-dimensional space form a basis, confusing the number of vectors with the dimension, and forgetting that infinite-dimensional spaces exist.
Concept
Start with two vector spaces over the same field, and a function from one to the other.
\[ T : V \to W \]
We call the input space the domain and the output space the codomain. Both are vector spaces over the same scalar field.
A vector space has exactly two operations: adding vectors, and scaling by a field element. A linear map is one that is compatible with both. That single idea is the whole topic.
Counterexample
Discussion prompt
Start with two vector spaces over the same field, and a function from one to the other.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
We call the input space the domain and the output space the codomain. Both are vector spaces over the same scalar field.
Concept
The first requirement: the map should not care whether you add first and then apply it, or apply it first and then add.
\[ T(u + v) = T(u) + T(v) \]
The plus sign on the left is addition in the domain; the plus on the right is addition in the codomain. The map carries one addition to the other.
Analogy
Discussion prompt
Explain Axiom 1: additivity by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
The first requirement: the map should not care whether you add first and then apply it, or apply it first and then add.
Concept
The second requirement: scaling a vector and then applying the map gives the same result as applying the map and then scaling.
\[ T(c\,v) = c\,T(v) \qquad \text{for every scalar } c \]
linear map — A function between vector spaces over the same field satisfying additivity and homogeneity. Also called a linear transformation or a vector-space homomorphism.
Explain it
Discussion prompt
Explain Axiom 2: homogeneity to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The second requirement: scaling a vector and then applying the map gives the same result as applying the map and then scaling.
Intuition
A vector space is not just a set of points; it is a set plus the ability to add and scale. A linear map is a function that carries that entire structure across intact.
Compare with earlier morphisms in this course: a group homomorphism respects the group operation, a ring homomorphism respects both ring operations. A linear map is the same idea for vector spaces.
Whenever you meet a new kind of object, the maps that respect its structure are where the real theory lives. For vector spaces, those maps are the linear ones.
Picture it
Figure (svg): A square grid on the left with the origin marked, and an arrow to a sheared grid on the right whose lines are still straight and parallel and whose origin is unmoved.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Picture an evenly ruled grid on the plane. A linear map may rotate, stretch, shear, or flatten that grid, but it keeps the lines straight, keeps parallel lines parallel, and pins the origin in place.
Intuition
Picture an evenly ruled grid on the plane. A linear map may rotate, stretch, shear, or flatten that grid, but it keeps the lines straight, keeps parallel lines parallel, and pins the origin in place.
Figure (svg): A square grid on the left with the origin marked, and an arrow to a sheared grid on the right whose lines are still straight and parallel and whose origin is unmoved.
Straight stays straight, parallel stays parallel, and the red origin never moves. That last fact is our next slide.
Concept
The two axioms combine into a single statement: a map is linear exactly when it preserves every finite linear combination.
\[ T\!\left(\sum_{i=1}^{n} c_i\, v_i\right) = \sum_{i=1}^{n} c_i\, T(v_i) \]
Additivity handles the sum, homogeneity handles each scalar, so checking this one equation for all combinations is equivalent to checking both axioms. In practice we usually verify the two axioms separately, since it is less to write.
Ranking
Put in order
Put the moves of Worked example: a map on the plane into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Apply T to a sum of two inputs and expand each output coordinate.
Worked example
Show that this map is linear.
\[ T:\mathbb{R}^2 \to \mathbb{R}^2,\qquad T(x,y) = (x+y,\; 2x) \]
Check additivity
Why: Apply T to a sum of two inputs and expand each output coordinate.
\[ T\big((x_1,y_1)+(x_2,y_2)\big) = \big((x_1+x_2)+(y_1+y_2),\; 2(x_1+x_2)\big) \]
Regroup into two separate outputs
Why: Rearranging the coordinates splits the result into T of the first input plus T of the second.
\[ = (x_1+y_1,\,2x_1) + (x_2+y_2,\,2x_2) = T(x_1,y_1) + T(x_2,y_2) \]
Check homogeneity
Why: Scale the input by c and factor c out of each output coordinate.
\[ T(c\,x,\,c\,y) = (cx+cy,\; 2cx) = c\,(x+y,\;2x) = c\,T(x,y) \]
Verify on concrete vectors
Why: Both axioms hold, so T is linear. As a spot check, T(1,0)=(1,2) and T(0,1)=(1,0) sum to (2,2), which equals T(1,1)=(2,2).
Picture it
Animation
Shows: Each line of the worked example "a map on the plane", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Apply T to a sum of two inputs and expand each output coordinate.
Concept
You never have to assume this separately; it is forced by homogeneity.
Take the scalar to be zero
Why: Homogeneity says T(c v) = c T(v) for every scalar; choose c equal to zero.
\[ T(\mathbf{0}) = T(0\cdot v) = 0\cdot T(v) = \mathbf{0} \]
So the zero vector of the domain always lands on the zero vector of the codomain. This gives a one-line rejection test for candidate maps.
Intuition
The origin is the one vector defined purely by the structure: it is the additive identity. A map that respects the structure has to send the identity of the domain to the identity of the codomain.
This is the same reason a group homomorphism sends identity to identity. Fixing the origin is not a special property of linear maps; it is a symptom of respecting the operations.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A tempting mistake: calling this map linear because its graph is a straight line.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: A linear map must fix zero, but this one does not.
The straight-line graph makes it affine, not linear. A line through the origin is linear; a shifted line is not.
Why: A linear map must fix zero, but this one does not.
Trap
A tempting mistake: calling this map linear because its graph is a straight line.
\[ f:\mathbb{R}\to\mathbb{R},\qquad f(x) = x + 1 \]
Test the origin
Why: A linear map must fix zero, but this one does not.
\[ f(0) = 0 + 1 = 1 \neq 0 \]
Confirm additivity also fails
Why: The failure is real, not a technicality: additivity breaks too.
\[ f(1)+f(1) = 4,\quad \text{but}\quad f(1+1) = f(2) = 3 \]
The straight-line graph makes it affine, not linear. A line through the origin is linear; a shifted line is not.
affine map — A linear map followed by a translation, of the form x maps to T(x) plus a fixed vector b. It is linear precisely when the shift b is zero.
Strip off the shift
Why: Removing the constant term restores the origin-fixing map, which is genuinely linear.
\[ g(x) = x \quad\Rightarrow\quad g(0)=0,\ \ g(x+y)=g(x)+g(y) \]
Definition probe
Sort into buckets
Every line below is part of the definition of linear map or of affine map — one or the other, never both. Put each where it belongs.
Concept
Once you have the two axioms, several facts follow with no extra assumptions.
\[ T(-v) = -T(v), \qquad T(u-v) = T(u) - T(v) \]
Negatives are just scaling by minus one, so homogeneity delivers these for free.
In short, a linear map commutes with every vector-space operation you can build from adding and scaling. That is exactly why the images of a basis will turn out to control the whole map.
Step zero
Discussion prompt
Worked example: differentiation is linear — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check additivity
Answer:
Worked example
The polynomials of degree at most three form a vector space; differentiation sends each to a polynomial of degree at most two.
\[ D : \mathcal{P}_3 \to \mathcal{P}_2,\qquad D(p) = p' \]
Check additivity
Why: The derivative of a sum is the sum of derivatives, a rule from calculus that holds termwise.
\[ D(p+q) = (p+q)' = p' + q' = D(p) + D(q) \]
Check homogeneity
Why: Constants pull out of a derivative, so scaling the input scales the output.
\[ D(c\,p) = (c\,p)' = c\,p' = c\,D(p) \]
Verify on a sample polynomial
Why: Both axioms hold, so D is linear. Check: with p equal to x cubed and q equal to x, D of their sum is 3x squared plus 1, matching D(p) plus D(q).
\[ D(x^3 + x) = 3x^2 + 1 = D(x^3) + D(x) \]
Picture it
Animation
Shows: Each line of the worked example "differentiation is linear", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The derivative of a sum is the sum of derivatives, a rule from calculus that holds termwise.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Another tempting map: squaring. It fixes the origin, so the quick test passes, and students often stop there.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Fixing the origin is necessary but not sufficient; additivity is where squaring breaks.
Passing the origin test proves nothing on its own. You must still check additivity and homogeneity; here the cross term is the culprit.
Why: Fixing the origin is necessary but not sufficient; additivity is where squaring breaks.
Trap
Another tempting map: squaring. It fixes the origin, so the quick test passes, and students often stop there.
\[ f:\mathbb{R}\to\mathbb{R},\qquad f(x) = x^2,\qquad f(0)=0 \]
Test additivity with real numbers
Why: Fixing the origin is necessary but not sufficient; additivity is where squaring breaks.
\[ f(1+2) = 9,\qquad f(1)+f(2) = 1 + 4 = 5,\qquad 9 \neq 5 \]
Passing the origin test proves nothing on its own. You must still check additivity and homogeneity; here the cross term is the culprit.
\[ (x+y)^2 = x^2 + 2xy + y^2 \neq x^2 + y^2 \]
Name the obstruction
Why: The extra 2xy term is exactly the gap between the map of a sum and the sum of the maps, so no scaling fixes it.
\[ f(x+y) - \big(f(x)+f(y)\big) = 2xy \]
Notation
Annotate
From Trap: squaring is not linear — read this one piece at a time. What is each part doing?
On: \( f:\mathbb{R}\to\mathbb{R},\qquad f(x) = x^2,\qquad f(0)=0 \)
Estimation
Predict first
Dropping the last coordinate of a three-dimensional vector gives a map to the plane.
Commit before you compute: what does Worked example: a projection is linear come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify on a concrete vector
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Both axioms hold, so P is linear.
Worked example
Dropping the last coordinate of a three-dimensional vector gives a map to the plane.
\[ P:\mathbb{R}^3 \to \mathbb{R}^2,\qquad P(x,y,z) = (x,\,y) \]
Check additivity
Why: Adding two inputs first, then dropping the third coordinate, matches dropping first and then adding.
\[ P\big((x_1,y_1,z_1)+(x_2,y_2,z_2)\big) = (x_1+x_2,\,y_1+y_2) = P(v_1)+P(v_2) \]
Check homogeneity
Why: Scaling every coordinate then discarding the third scales the surviving two.
\[ P(c\,x,\,c\,y,\,c\,z) = (cx,\,cy) = c\,P(x,y,z) \]
Verify on a concrete vector
Why: Both axioms hold, so P is linear. Check: P(2,3,9) is (2,3), and P(4,6,18) is (4,6), which is twice (2,3) as homogeneity predicts.
Picture it
Animation
Shows: Each line of the worked example "a projection is linear", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Adding two inputs first, then dropping the third coordinate, matches dropping first and then adding.
Concept
Here is the fact that makes linear maps so manageable: you do not need to know a linear map on every vector. You only need to know it on a basis.
If the domain has a basis and you know where the map sends each basis vector, then the value on any other vector is forced.
\[ v = c_1 b_1 + \cdots + c_n b_n \ \Rightarrow\ T(v) = c_1 T(b_1) + \cdots + c_n T(b_n) \]
The coordinates are unique because a basis gives a unique representation, so the output is unambiguous. Knowing finitely many outputs pins down infinitely many.
Intuition
Think of the basis vectors as the skeleton of the space. Every other vector is a specific recipe of basis vectors, and linearity says the map treats each vector by the same recipe applied to the outputs.
This is why a two-dimensional linear map is captured by just two output vectors, and an n-dimensional one by n outputs. It is the seed of the entire matrix idea in the next deck: the columns of a matrix are exactly the images of the basis vectors.
Concept
The determination fact has a converse, and together they form a clean existence-and-uniqueness statement.
Pick any basis of the domain and any vectors you like in the codomain as their intended images. There is exactly one linear map achieving those images.
\[ T(b_i) = w_i \ \text{(chosen freely)} \ \Rightarrow\ \exists!\ \text{linear } T \text{ with this action} \]
This is the universal property of a basis: a basis is a set of inputs you can map anywhere, with the map extending uniquely. Freedom to choose the images, and uniqueness once you choose, live side by side.
Missing information
Discussion prompt
Define a linear map on the plane by choosing where the standard basis vectors go.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Every vector in the plane is a combination of the standard basis vectors with its own coordinates.
Worked example
Define a linear map on the plane by choosing where the standard basis vectors go.
\[ T(e_1) = (2,1),\qquad T(e_2) = (0,3) \]
Write a general vector in the basis
Why: Every vector in the plane is a combination of the standard basis vectors with its own coordinates.
\[ (x,y) = x\,e_1 + y\,e_2 \]
Apply linearity
Why: Push T through the combination, replacing each basis vector by its chosen image.
\[ T(x,y) = x\,T(e_1) + y\,T(e_2) = x(2,1) + y(0,3) = (2x,\; x+3y) \]
Verify the basis images come back
Why: Plug in the basis vectors: T(1,0) is (2,1) and T(0,1) is (0,3), matching the chosen images, so the formula is the unique linear extension.
Picture it
Animation
Shows: Each line of the worked example "build a map from its basis action", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Plug in the basis vectors: T(1,0) is (2,1) and T(0,1) is (0,3), matching the chosen images, so the formula is the unique linear extension.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Suppose you are told only one output and asked for the whole map.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: a single vector is not a basis of the plane, so it cannot fix the map.
You need images on a whole basis. One vector spans only a line, leaving every direction off that line undetermined.
Why: a single vector is not a basis of the plane, so it cannot fix the map.
Trap
Suppose you are told only one output and asked for the whole map.
\[ T(1,1) = (2,2)\ \Rightarrow\ T = \,? \]
Claim the map is now determined
Why: This is the error: a single vector is not a basis of the plane, so it cannot fix the map.
Two genuinely different linear maps both send this vector to the same place.
\[ S(x,y) = (x+y,\,x+y),\qquad R(x,y) = (2x,\,2y) \]
\[ S(1,1) = (2,2) = R(1,1), \quad\text{yet}\quad S(1,0)=(1,1)\neq(2,0)=R(1,0) \]
You need images on a whole basis. One vector spans only a line, leaving every direction off that line undetermined.
\[ \dim \mathbb{R}^2 = 2 \ \Rightarrow\ \text{need 2 basis images} \]
Supply a second independent image
Why: Once you also fix T on a second vector completing a basis, the extension theorem makes the map unique.
\[ T(1,1) = (2,2),\ T(1,-1) = w \ \Rightarrow\ T \text{ determined} \]
Translation
\( S(1,1) = (2,2) = R(1,1), \quad\text{yet}\quad S(1,0)=(1,1)\neq(2,0)=R(1,0) \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Worked example
Two linear maps that give the same output on every basis vector are the same map. This is the determination fact turned into a proof technique.
\[ S(b_i) = T(b_i) \ \text{for all } i \ \Rightarrow\ S = T \]
Take an arbitrary vector and expand it
Why: To show two functions are equal, show they agree on every input; write the input in the basis.
\[ v = \textstyle\sum_i c_i b_i \]
Push both maps through the combination
Why: Linearity lets each map act coordinatewise, and the basis outputs are assumed equal.
\[ S(v) = \textstyle\sum_i c_i S(b_i) = \sum_i c_i T(b_i) = T(v) \]
Verify the conclusion holds for every v
Why: Since v was arbitrary and S(v) equals T(v) in all cases, the two maps are identical as functions.
Picture it
Animation
Shows: Each line of the worked example "agreeing on a basis means equal", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Since v was arbitrary and S(v) equals T(v) in all cases, the two maps are identical as functions.
Concept
Now we study the two subspaces that measure how a linear map behaves. The first lives in the domain.
\[ \ker T = \{\, v \in V : T(v) = \mathbf{0} \,\} \]
kernel — The set of all domain vectors that the map sends to the zero vector of the codomain. Also called the null space of T.
The kernel always contains at least the zero vector, since every linear map fixes the origin. How much more it contains is the whole question.
Intuition
Think of the kernel as the collection of directions the map flattens to nothing. A projection that forgets the vertical coordinate crushes the entire vertical axis to zero; that axis is its kernel.
The bigger the kernel, the more information the map throws away, and the further it is from being reversible. A trivial kernel means nothing is lost, which is precisely injectivity, as we will prove shortly.
Step zero
Discussion prompt
Worked example: computing a kernel — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set the output equal to zero
Answer:
Worked example
Find the kernel of this map from three-dimensional space to the plane.
\[ T:\mathbb{R}^3 \to \mathbb{R}^2,\qquad T(x,y,z) = (x-y,\; y-z) \]
Set the output equal to zero
Why: A vector is in the kernel exactly when both output coordinates vanish, giving two equations.
\[ x - y = 0, \qquad y - z = 0 \]
Solve the system
Why: The equations force all three coordinates to be equal, so the solutions form a single line of vectors.
\[ x = y = z \ \Rightarrow\ (x,y,z) = t\,(1,1,1) \]
Verify the spanning vector is in the kernel
Why: Check the witness: T(1,1,1) equals (1 minus 1, 1 minus 1), which is (0,0), so the kernel is the line spanned by (1,1,1).
\[ \ker T = \operatorname{span}\{(1,1,1)\} \]
Picture it
Animation
Shows: Each line of the worked example "computing a kernel", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check the witness: T(1,1,1) equals (1 minus 1, 1 minus 1), which is (0,0), so the kernel is the line spanned by (1,1,1).
Concept
The second subspace lives in the codomain: it is everything the map actually reaches.
\[ \operatorname{im} T = \{\, T(v) : v \in V \,\} \]
image — The set of all outputs the map produces, as the input ranges over the whole domain. Also called the range of T.
Because a basis determines the map, the image is spanned by the images of the basis vectors. To describe the image, apply the map to a basis and take the span.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of linear map, affine map, kernel, image as Linear Maps uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Estimation
Predict first
Return to differentiation on polynomials of degree at most three, mapping into degree at most two.
Commit before you compute: what does Worked example: kernel and image of differentiation come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify surjectivity with a witness
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Check a target: the polynomial x squared is D of one-third x cubed, and any degree-two polynomial has an antiderivative in the domain, so the image is all of the codomain.
Worked example
Return to differentiation on polynomials of degree at most three, mapping into degree at most two.
\[ D:\mathcal{P}_3 \to \mathcal{P}_2,\qquad D(p) = p' \]
Find the kernel
Why: A polynomial has zero derivative exactly when it has no varying terms, that is, when it is constant.
\[ \ker D = \{\, p : p' = 0 \,\} = \{\, \text{constant polynomials} \,\} \]
Find the image
Why: Every polynomial of degree at most two is the derivative of some polynomial of degree at most three, obtained by antidifferentiating.
\[ \operatorname{im} D = \mathcal{P}_2 \]
Verify surjectivity with a witness
Why: Check a target: the polynomial x squared is D of one-third x cubed, and any degree-two polynomial has an antiderivative in the domain, so the image is all of the codomain.
\[ D\!\left(\tfrac{1}{3}x^3\right) = x^2 \]
Picture it
Animation
Shows: Each line of the worked example "kernel and image of differentiation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check a target: the polynomial x squared is D of one-third x cubed, and any degree-two polynomial has an antiderivative in the domain, so the image is all of the codomain.
Concept
The kernel is never just a random subset: it is always a subspace of the domain.
That means it passes the subspace test: it contains the zero vector, and it is closed under both addition and scalar multiplication. We prove this next, and it is a template you will reuse constantly.
Ranking
Put in order
Put the moves of Worked example: proving the kernel is a subspace into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Every linear map fixes the origin, so the zero vector maps to zero and lies in the kernel; the kernel is nonempty.
Worked example
Show the kernel satisfies the three-part subspace test.
Contains the zero vector
Why: Every linear map fixes the origin, so the zero vector maps to zero and lies in the kernel; the kernel is nonempty.
\[ T(\mathbf{0}) = \mathbf{0} \ \Rightarrow\ \mathbf{0} \in \ker T \]
Closed under addition
Why: If two vectors both map to zero, additivity sends their sum to zero plus zero.
\[ u,v \in \ker T \ \Rightarrow\ T(u+v) = T(u)+T(v) = \mathbf{0} \]
Closed under scaling
Why: Scaling a kernel vector by any scalar keeps it in the kernel, by homogeneity.
\[ v \in \ker T \ \Rightarrow\ T(cv) = c\,T(v) = c\,\mathbf{0} = \mathbf{0} \]
Verify all three conditions hold
Why: The kernel contains zero and is closed under addition and scaling, so by the subspace test it is a subspace of the domain.
Picture it
Animation
Shows: Each line of the worked example "proving the kernel is a subspace", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The kernel contains zero and is closed under addition and scaling, so by the subspace test it is a subspace of the domain.
Concept
The image is a subspace of the codomain, by a parallel argument. The difference is that its vectors are described as outputs, so closure is checked by pulling back to inputs.
This symmetry, kernel in the domain and image in the codomain, is worth holding onto; confusing the two is a classic error we will flag explicitly.
Missing information
Discussion prompt
Show the image satisfies the subspace test inside the codomain.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The zero of the codomain is the output of the zero of the domain, so it lies in the image.
Worked example
Show the image satisfies the subspace test inside the codomain.
Contains the zero vector
Why: The zero of the codomain is the output of the zero of the domain, so it lies in the image.
\[ T(\mathbf{0}) = \mathbf{0} \ \Rightarrow\ \mathbf{0} \in \operatorname{im} T \]
Closed under addition
Why: Two image vectors are outputs of some inputs; the output of the sum of those inputs is their sum, again an output.
\[ T(u) + T(v) = T(u+v) \in \operatorname{im} T \]
Closed under scaling
Why: A scalar times an output is the output of that scalar times the input, so it stays in the image.
\[ c\,T(v) = T(cv) \in \operatorname{im} T \]
Verify all three conditions hold
Why: The image contains zero and is closed under addition and scaling, so it is a subspace of the codomain.
Intuition
Keep a clear picture of where each object lives. The kernel sits inside the domain; the image sits inside the codomain. They measure different things: how much the map collapses, and how much it covers.
Figure (svg): Two ovals labeled domain V and codomain W. Inside V a smaller region is shaded and labeled kernel. An arrow labeled T goes from V to W, and inside W a smaller region is shaded and labeled image.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A subtle mistake: writing the kernel as a subset of the codomain, or confusing it with the zero of the codomain.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The kernel is made of inputs. Inputs live in the domain, so the kernel cannot sit inside the codomain at all.
The kernel is a subspace of the domain; the image is a subspace of the codomain. The zero vector in the kernel is the domain's zero, and the target it maps to is the codomain's zero.
Why: The kernel is made of inputs. Inputs live in the domain, so the kernel cannot sit inside the codomain at all.
Trap
A subtle mistake: writing the kernel as a subset of the codomain, or confusing it with the zero of the codomain.
\[ T:\mathbb{R}^3\to\mathbb{R}^2,\qquad \text{``}\,\ker T = \{\mathbf{0}\}\subseteq \mathbb{R}^2\,\text{''} \]
Spot the type error
Why: The kernel is made of inputs. Inputs live in the domain, so the kernel cannot sit inside the codomain at all.
The condition defining the kernel is a statement about which inputs vanish, not about the output space.
The kernel is a subspace of the domain; the image is a subspace of the codomain. The zero vector in the kernel is the domain's zero, and the target it maps to is the codomain's zero.
\[ \ker T \subseteq V \ (=\mathbb{R}^3), \qquad \operatorname{im} T \subseteq W \ (=\mathbb{R}^2) \]
Read the definition literally
Why: The defining condition ranges over v in V, so the set it carves out is inside V, the domain.
\[ \ker T = \{\, v \in V : T(v) = \mathbf{0}_W \,\} \]
Notation
Annotate
From Trap: the kernel lives in the wrong space — read this one piece at a time. What is each part doing?
On: \( \ker T = \{\, v \in V : T(v) = \mathbf{0}_W \,\} \)
Concept
The kernel does more than measure collapse; it decides injectivity outright, with the cleanest possible test.
\[ T \text{ is injective} \iff \ker T = \{\mathbf{0}\} \]
So to check whether a linear map is one-to-one, you never chase pairs of inputs. You compute a single set, the kernel, and ask whether it is just the zero vector. This is the reason kernels are so useful.
Intuition
Two inputs collide when they share an output. For a linear map, that happens exactly when their difference is crushed to zero, because the map of a difference is the difference of the outputs.
So every collision is packaged as a nonzero kernel vector, and every nonzero kernel vector creates a collision with the origin. No nonzero kernel vectors means no collisions, which is injectivity. That single equivalence is the content of the next proof.
Step zero
Discussion prompt
Worked example: injective if and only if trivial kernel — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Forward: injective forces a trivial kernel
Answer:
Worked example
Prove both directions of the equivalence for a linear map T.
Forward: injective forces a trivial kernel
Why: Take any kernel vector; it maps to zero, and so does the origin, so injectivity makes them equal.
\[ v \in \ker T \Rightarrow T(v) = \mathbf{0} = T(\mathbf{0}) \ \overset{\text{inj}}{\Rightarrow}\ v = \mathbf{0} \]
Backward: trivial kernel forces injectivity
Why: Suppose two inputs share an output; subtract to send their difference to zero, so the difference lies in the kernel.
\[ T(u) = T(v) \Rightarrow T(u-v) = \mathbf{0} \Rightarrow u - v \in \ker T \]
Use triviality of the kernel
Why: If the only kernel vector is zero, the difference must be zero, so the two inputs coincide.
\[ \ker T = \{\mathbf{0}\} \Rightarrow u - v = \mathbf{0} \Rightarrow u = v \]
Verify both implications are established
Why: Injectivity gives a trivial kernel, and a trivial kernel gives injectivity, so the equivalence holds in both directions.
Picture it
Animation
Shows: Each line of the worked example "injective if and only if trivial kernel", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Injectivity gives a trivial kernel, and a trivial kernel gives injectivity, so the equivalence holds in both directions.
Concept
The companion notion is surjectivity, and it is read off the image just as directly.
\[ T \text{ is surjective} \iff \operatorname{im} T = W \]
Injectivity is a kernel question in the domain; surjectivity is an image question in the codomain. Keeping these on the correct side of the map prevents most of the errors in this topic.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A strong-student reflex from finite dimensions: assuming an injective linear map must also be surjective. In infinite dimensions this fails.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The only sequence sent to all zeros is the all-zero sequence, so the kernel is trivial and S is injective.
Yet the shift is not surjective: no sequence maps to one with a nonzero first entry, because the first output slot is always zero.
Why: The only sequence sent to all zeros is the all-zero sequence, so the kernel is trivial and S is injective.
Trap
A strong-student reflex from finite dimensions: assuming an injective linear map must also be surjective. In infinite dimensions this fails.
Consider the right-shift map on infinite sequences of real numbers.
\[ S(a_1,a_2,a_3,\dots) = (0,a_1,a_2,a_3,\dots) \]
Note it is injective
Why: The only sequence sent to all zeros is the all-zero sequence, so the kernel is trivial and S is injective.
\[ \ker S = \{\mathbf{0}\} \]
Yet the shift is not surjective: no sequence maps to one with a nonzero first entry, because the first output slot is always zero.
\[ (1,0,0,\dots) \notin \operatorname{im} S \]
State the correct hypothesis
Why: The injective-implies-surjective miracle needs a linear map from a finite-dimensional space to one of the same dimension; infinite-dimensional spaces escape it.
For an endomorphism of a finite-dimensional space, injective, surjective, and bijective do coincide, and the rank-nullity theorem in the next deck explains exactly why.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
A cornerstone geometric example: rotating the plane about the origin by a fixed angle.
\[ R_\theta(x,y) = (x\cos\theta - y\sin\theta,\ \ x\sin\theta + y\cos\theta) \]
Each output coordinate is a fixed combination of the inputs with constant coefficients, and any map whose outputs are constant-coefficient combinations of the inputs is linear. We verify this and find its kernel next.
Fill the middle
Fill in the blanks
From Worked example: rotation is linear with trivial kernel — finish the line. Write what belongs on the right of the equals sign before you look.
R_\theta(x,y) = (x\cos\theta - y\sin\theta,\ x\sin\theta + y\cos\theta)
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Scaling both inputs by c pulls c out of every term of both output coordinates.
Worked example
Verify linearity of rotation and determine its kernel.
\[ R_\theta(x,y) = (x\cos\theta - y\sin\theta,\ x\sin\theta + y\cos\theta) \]
Check homogeneity
Why: Scaling both inputs by c pulls c out of every term of both output coordinates.
\[ R_\theta(cx,cy) = c\,(x\cos\theta - y\sin\theta,\ x\sin\theta + y\cos\theta) = c\,R_\theta(x,y) \]
Check additivity
Why: Because each coordinate is a sum of terms linear in x and y, splitting the inputs splits the outputs.
\[ R_\theta(u+v) = R_\theta(u) + R_\theta(v) \]
Find the kernel
Why: Rotation preserves length, so the only vector sent to the origin is the origin itself; the kernel is trivial and rotation is injective.
\[ R_\theta(v) = \mathbf{0} \Rightarrow \|v\| = \|R_\theta(v)\| = 0 \Rightarrow v = \mathbf{0} \]
Verify with a quarter turn
Why: Check a case: at a quarter turn, cosine is 0 and sine is 1, so the map sends (1,0) to (0,1), a genuine rotation that moves no nonzero vector to the origin.
Picture it
Animation
Shows: Each line of the worked example "rotation is linear with trivial kernel", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Scaling both inputs by c pulls c out of every term of both output coordinates.
Hypothesis
Predict first
Worked example: a map into a bigger space is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Confirm linearity by structure
Why: Every output coordinate is a constant-coefficient combination of x and y, so additivity and homogeneity both hold.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Linear maps can raise dimension. Consider this map from the plane into three-dimensional space.
\[ T:\mathbb{R}^2 \to \mathbb{R}^3,\qquad T(x,y) = (x - y,\ y - x,\ x + y) \]
Confirm linearity by structure
Why: Every output coordinate is a constant-coefficient combination of x and y, so additivity and homogeneity both hold.
\[ T(c\,u + d\,v) = c\,T(u) + d\,T(v) \]
Solve for the kernel
Why: Set every output coordinate to zero; the first two give x equal to y, and the third then forces both to vanish.
\[ x - y = 0,\ \ x + y = 0 \ \Rightarrow\ x = y = 0 \]
Verify the kernel is trivial
Why: Only the zero vector solves the system, so the kernel is just the origin and T is injective; check that T(0,0) is indeed (0,0,0).
\[ \ker T = \{\mathbf{0}\} \]
Picture it
Animation
Shows: Each line of the worked example "a map into a bigger space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Only the zero vector solves the system, so the kernel is just the origin and T is injective; check that T(0,0) is indeed (0,0,0).
Concept
Collect all the linear maps between two fixed vector spaces into one set. That set is itself a vector space.
\[ \mathcal{L}(V,W) = \{\, T : V \to W \ \text{linear} \,\} \]
You add two maps by adding their outputs, and scale a map by scaling its outputs. These operations are defined pointwise.
\[ (S+T)(v) = S(v)+T(v), \qquad (c\,T)(v) = c\,T(v) \]
Intuition
The point of bundling maps into a space is that the tools of this whole unit now apply to maps themselves. You can ask for a basis of the space of maps, its dimension, and its subspaces.
One checks that the pointwise sum and scalar multiple of linear maps are again linear, and the zero map is the additive identity. Every vector-space axiom is inherited from the codomain, so nothing new needs to be verified from scratch.
Concept
When the codomain of one map is the domain of another, you can chain them, and the chain is again linear.
\[ T:V\to W,\quad S:W\to U \ \Rightarrow\ S\circ T : V \to U \]
This is the fact that will make matrix multiplication mean composition in the next deck. Composition is associative and distributes over addition of maps, though it need not be commutative.
Estimation
Predict first
Prove that composing two linear maps yields a linear map.
Commit before you compute: what does Worked example: a composition is linear come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify both axioms hold for the composite
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The composite is additive and homogeneous, so it is linear; each move used only the linearity of S or of T.
Worked example
Prove that composing two linear maps yields a linear map.
\[ S:W\to U,\quad T:V\to W \text{ both linear} \]
Check additivity of the composite
Why: Feed a sum into T, use T additive, then use S additive on the result.
\[ (S\circ T)(u+v) = S\big(T(u)+T(v)\big) = S(T(u)) + S(T(v)) \]
Check homogeneity of the composite
Why: Pull the scalar through T first, then through S; each step is justified by one map's homogeneity.
\[ (S\circ T)(cv) = S\big(c\,T(v)\big) = c\,S(T(v)) = c\,(S\circ T)(v) \]
Verify both axioms hold for the composite
Why: The composite is additive and homogeneous, so it is linear; each move used only the linearity of S or of T.
Picture it
Animation
Shows: Each line of the worked example "a composition is linear", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Feed a sum into T, use T additive, then use S additive on the result.
Concept
When a linear map goes from a space back to itself, it is called an endomorphism, or a linear operator.
\[ \operatorname{End}(V) = \mathcal{L}(V,V) \]
These maps can be added, scaled, and composed, all staying inside the same set. The composition acts as a multiplication.
That combination of a vector space plus a compatible multiplication makes the operators an algebra. The identity map is the multiplicative identity, echoing the monoid of functions you saw earlier in the course.
Concept
The space of all infinite real sequences is a vector space, and shifting is a natural operator on it. We met the right shift as a trap; here we study it as a genuine example.
\[ S(a_1,a_2,a_3,\dots) = (0,a_1,a_2,\dots) \]
This map is important precisely because it is injective without being surjective, which is impossible for a self-map of a finite-dimensional space. Infinite dimensions are where such phenomena live.
Explain it
Discussion prompt
Explain The shift operator on sequences to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The space of all infinite real sequences is a vector space, and shifting is a natural operator on it. We met the right shift as a trap; here we study it as a genuine example.
Step zero
Discussion prompt
Worked example: the shift is linear and injective — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check additivity
Answer:
Worked example
Verify the right shift is linear and find its kernel.
\[ S(a_1,a_2,\dots) = (0,a_1,a_2,\dots) \]
Check additivity
Why: Shifting the sum of two sequences produces the same slots as summing the two shifted sequences, entry by entry.
\[ S(a+b) = (0,a_1+b_1,\dots) = S(a) + S(b) \]
Check homogeneity
Why: Scaling every entry then shifting equals shifting then scaling, since the inserted zero is unaffected by scaling.
\[ S(c\,a) = (0,c\,a_1,c\,a_2,\dots) = c\,S(a) \]
Find the kernel
Why: If the shifted sequence is all zeros, then every original entry is zero, so only the zero sequence is in the kernel.
\[ S(a) = \mathbf{0} \Rightarrow a_1 = a_2 = \cdots = 0 \Rightarrow \ker S = \{\mathbf{0}\} \]
Verify injectivity via the kernel
Why: The kernel is trivial, so by our equivalence the shift is injective, even though it fails to be surjective.
Picture it
Animation
Shows: Each line of the worked example "the shift is linear and injective", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Shifting the sum of two sequences produces the same slots as summing the two shifted sequences, entry by entry.
Concept
A linear map whose codomain is the scalar field itself is called a linear functional. A basic family of them evaluates a polynomial at a fixed point.
\[ \operatorname{ev}_a : \mathcal{P}_n \to \mathbb{R},\qquad \operatorname{ev}_a(p) = p(a) \]
Fix the point once; the input is the polynomial, and the output is the number you get by plugging in.
Functionals are the objects of the dual space, which the invertibility-and-duality deck develops. For now they are simply a clean source of linear maps into the scalars.
Analogy
Discussion prompt
Explain Evaluation functionals by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A linear map whose codomain is the scalar field itself is called a linear functional. A basic family of them evaluates a polynomial at a fixed point.
Ranking
Put in order
Put the moves of Worked example: evaluation is linear into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The sum of two polynomials, evaluated at a, is the sum of their values, by the definition of adding functions.
Worked example
Show that evaluating at a fixed point is a linear functional.
\[ \operatorname{ev}_a(p) = p(a) \]
Check additivity
Why: The sum of two polynomials, evaluated at a, is the sum of their values, by the definition of adding functions.
\[ \operatorname{ev}_a(p+q) = (p+q)(a) = p(a) + q(a) = \operatorname{ev}_a(p) + \operatorname{ev}_a(q) \]
Check homogeneity
Why: A scalar multiple of a polynomial, evaluated at a, scales the value by the same scalar.
\[ \operatorname{ev}_a(c\,p) = (c\,p)(a) = c\,p(a) = c\,\operatorname{ev}_a(p) \]
Verify with a concrete evaluation
Why: Check at the point 2 with the squaring polynomial: its value is 4, and doubling the polynomial gives value 8, exactly twice, as homogeneity requires.
\[ \operatorname{ev}_2(x^2) = 4, \qquad \operatorname{ev}_2(2x^2) = 8 \]
Picture it
Animation
Shows: Each line of the worked example "evaluation is linear", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sum of two polynomials, evaluated at a, is the sum of their values, by the definition of adding functions.
Concept
Two extreme operators anchor the whole picture. The identity map changes nothing; the zero map destroys everything.
\[ \operatorname{id}(v) = v, \qquad \ker(\operatorname{id}) = \{\mathbf{0}\},\ \ \operatorname{im}(\operatorname{id}) = V \]
\[ Z(v) = \mathbf{0}, \qquad \ker Z = V,\ \ \operatorname{im} Z = \{\mathbf{0}\} \]
The identity has the smallest possible kernel and largest possible image; the zero map is its exact opposite. Every other operator sits between these poles.
Counterexample
Discussion prompt
Two extreme operators anchor the whole picture. The identity map changes nothing; the zero map destroys everything.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The identity has the smallest possible kernel and largest possible image; the zero map is its exact opposite. Every other operator sits between these poles.
Pattern
1. Try the origin first
Why: If the map fails to send the zero vector to the zero vector, it is not linear and you can stop immediately.
2. Check additivity
Why: Apply the map to a sum of two general inputs and confirm it equals the sum of the two outputs.
3. Check homogeneity
Why: Apply the map to a general scalar times an input and confirm the scalar factors out of the output.
Shortcut for coordinate formulas
Why: If every output coordinate is a constant-coefficient combination of the inputs with no constants and no products or powers, the map is automatically linear.
Pattern
Kernel: set the output to zero
Why: Write the equation that the output is the zero vector and solve the resulting homogeneous system; the solution set is the kernel, living in the domain.
Image: apply the map to a basis
Why: The image is the span of the images of a basis of the domain, because a basis determines the whole map; the image lives in the codomain.
Sanity-check the homes
Why: Confirm the kernel is a subset of the domain and the image is a subset of the codomain before quoting either one.
Pattern
1. Compute the kernel
Why: Injectivity for a linear map is governed entirely by the kernel, so never chase pairs of inputs directly.
2. Ask whether the kernel is trivial
Why: The map is injective exactly when the kernel contains only the zero vector; a single nonzero kernel vector kills injectivity.
3. Do not conflate with surjectivity
Why: A trivial kernel gives injectivity only; surjectivity is a separate image question, and the two coincide only for self-maps of a finite-dimensional space.
Pattern
1. Choose images for the basis vectors
Why: Pick any target vectors you like for each basis vector; the extension theorem guarantees a unique linear map with that action.
2. Expand a general input in the basis
Why: Write the general vector in terms of its unique coordinates against the basis.
3. Apply linearity to get the formula
Why: Replace each basis vector by its chosen image and combine with the same coordinates to read off the formula for the whole map.
Real world
Discussion prompt
Outside this lesson: where does Linear Maps actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: building a map from a basis is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck presents linear maps as the structure-preserving morphisms between vector spaces. It gives the two axioms, explains why the origin is fixed, shows that a basis determines the whole map, and covers the kernel and image as subspaces and injectivity as the kernel being trivial. It targets the classic traps of calling affine or squaring maps linear, thinking that one vector pins a map down, locating the kernel in the wrong space, and assuming that injective forces surjective.
Elimination
Eliminate the wrong options
Which of these maps from the plane to the plane is linear?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Only the first map has each output coordinate as a constant-coefficient combination of the inputs, with no constants, products, or powers, so it satisfies additivity and homogeneity. Each other option breaks a linearity axiom in a specific way.
Check
Exactly one of these four maps on the plane is linear. Test each against the axioms before choosing.
Check your understanding
Which of these maps from the plane to the plane is linear?
Answer: A
Why: Only the first map has each output coordinate as a constant-coefficient combination of the inputs, with no constants, products, or powers, so it satisfies additivity and homogeneity. Each other option breaks a linearity axiom in a specific way.
Prediction
Predict first
The map f(x) = 2x + 1 is not linear. Which single fact proves it most directly?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: f(0) = 1, but a linear map must send 0 to 0.
Why: Homogeneity forces every linear map to fix the origin, so a map with f(0) equal to 1 rather than 0 cannot be linear. Checking the origin is the fastest disqualifying test.
Check
Consider the map on the real line given by doubling and then adding one. It is not linear.
Check your understanding
The map f(x) = 2x + 1 is not linear. Which single fact proves it most directly?
Answer: A
Why: Homogeneity forces every linear map to fix the origin, so a map with f(0) equal to 1 rather than 0 cannot be linear. Checking the origin is the fastest disqualifying test.
Prediction
Predict first
What is the kernel of this differentiation map?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: The constant polynomials.
Why: A polynomial has zero derivative exactly when it is constant, so the kernel is the set of constant polynomials, a one-dimensional subspace of the domain.
Check
Take differentiation from the polynomials of degree at most three into the polynomials of degree at most two.
Check your understanding
What is the kernel of this differentiation map?
Answer: A
Why: A polynomial has zero derivative exactly when it is constant, so the kernel is the set of constant polynomials, a one-dimensional subspace of the domain.
Check
A linear map on the plane is specified by its action on the standard basis.
\[ T(1,0) = (3,1),\qquad T(0,1) = (-1,2) \]
Check your understanding
Using linearity, what is T(2, 5)?
Answer: A
Why: By linearity T(2,5) equals 2 times T(1,0) plus 5 times T(0,1), which is 2(3,1) plus 5(-1,2), giving (6-5, 2+10) equal to (1,12).
Elimination
Eliminate the wrong options
What follows immediately from a trivial kernel?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: A linear map is injective exactly when its kernel is trivial, so a kernel equal to just the zero vector makes T one-to-one on its domain.
Check
Suppose a linear map has a kernel consisting of only the zero vector.
\[ \ker T = \{\mathbf{0}\} \]
Check your understanding
What follows immediately from a trivial kernel?
Answer: A
Why: A linear map is injective exactly when its kernel is trivial, so a kernel equal to just the zero vector makes T one-to-one on its domain.
Check
Recall the right-shift operator on infinite sequences, which inserts a zero at the front. It has a trivial kernel, so it is injective.
\[ S(a_1,a_2,\dots) = (0,a_1,a_2,\dots) \]
Check your understanding
Is the right-shift operator surjective?
Answer: A
Why: The first output slot of a shifted sequence is always zero, so a sequence such as (1,0,0,...) is never produced; the shift is injective but not onto its codomain.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Pattern: testing a map for linearity · Pattern: finding the kernel and image · Pattern: deciding injectivity · Pattern: building a map from a basis · A linear map, informally. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A linear map is a function between vector spaces that respects addition and scalar multiplication. Those two axioms are the entire definition, and everything else is a consequence.
\[ T(u+v) = T(u)+T(v), \qquad T(cv) = c\,T(v) \]
The origin is always fixed, which rejects affine maps at a glance, and a linear map is completely determined by its values on a basis, which is the seed of matrices in the next deck.
The kernel lives in the domain and measures collapse; the image lives in the codomain and measures coverage. Both are subspaces, and injectivity is exactly a trivial kernel.
| Object | Where it lives | What it detects |
|---|---|---|
| Kernel | Domain | Injectivity (trivial means one-to-one) |
| Image | Codomain | Surjectivity (all of W means onto) |
Next comes the rank-nullity theorem, which links the sizes of the kernel and image and explains why, in finite dimensions, injective and surjective become the same condition.
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